Question

a number is selected randomly from the Integers 1 to 100 what is the probability that it will be a multiple of 4 or 6​

Answer 1

Answer:

33 out of 100

Step-by-step explanation:

There are 100 numbers in integers 1 to 100

ie S = (1,2,3....10)

n(s) = 100

Let Event $E_{1}$ be the numbers which are multiples of 4

$E_{1}$ = (x4) = [{$\frac{100}{4}$}] = [25] = 25

Therefore Probability of number which is a multiple of 4

= $P(E_{1}) = \frac{25}{100}$

Let Event $E_{2}$ be the numbers which are multiples of 6

$E_{2}$ = (x6) = [{$\frac{100}{6}$}] = [16.666] = 16

Therefore Probability of number which is a multiple of 6

= $P(E_{2}) = \frac{16}{100}$

Now we have to find the numbers in an Event where the number can be a multiple of both 4 and 6

For that we have to find the LCM of 4 and 6 which is 12

So let the Event be $E_{1} n E_{2}$

$E_{1} n E_{2}$ =( x12) =  [{$\frac{100}{12}$}] = [8.33333] = 8

Therefore Probability of number which is a multiple of both 4 and 6

= $P(E_{2} n E_{2} ) = \frac{8}{100}$

Now we have to find the probability that the number picked will be a multiple of 4 and 6

is $P(E_{1} u E_{2} )$ = $P(E_{1}) + P(E_{2}) - P(E_{1} n E_{2})$

= $\frac{25}{100} + \frac{16}{100} - \frac{8}{100}$

= $\frac{33}{100}$

Therefore the probability of a number which is selected from the integers 1 to 100 which is a multiple of 4 or 6 is 33 out of 100

Answer 2

Answer:

Step-by-step explanation:

Total cases = 100

Let A be the event which is to be a multiples of 4 and B be the event which is to be a multiples of 6

Hence, P(A) = 25/100

P(B) = 16/100

P(A∩B)= 8/100

Hence, the probability that a chosen number is a multiple of 4 or 6, P(AUB) = P(A) + P(B) – P(A∩B)

P(AUB) = (25/100) + (16/100) – (8/100)

P(AUB) = (25+16-8)/100

P(AUB) = 33/100.